Conditions for Losslessness

The scattering matrices for lossless physical waveguide junctions give an apparently unexplored class of lossless FDN prototypes. However, this is just a subset of all possible lossless feedback matrices. We are therefore interested in the most general conditions for losslessness of an FDN feedback matrix.

Consider the general case in which ${\bf A}$ is allowed to be any scattering matrix, i.e., it is associated with a not-necessarily-physical junction of $N$ physical waveguides. Following the definition of losslessness in classical network theory, we may say that a waveguide scattering matrix $A$ is said to be lossless if the total complex power [1] at the junction is scattering invariant, i.e.,

$${\bf {p}^+}^\ast {\bf\Gamma}\bf {p}^+ = {\bf {p}^-}^\ast {\bf\Gamma}\bf {p}^-$$

$$\Rightarrow \quad {\bf A}^\ast {\bf\Gamma}{\bf A} = {\bf\Gamma}$$ (27)

where ${\bf\Gamma}$ is any Hermitian, positive-definite matrix (which has an interpretation as a generalized junction admittance). The form $x^\ast {\bf\Gamma} x$ is by definition the square of the elliptic norm of $x$ induced by ${\bf\Gamma}$, or $\vert\vert x\vert\vert _{\bf\Gamma}^2 = x^\ast {\bf\Gamma}x$. Setting ${\bf\Gamma}={\bf I}$, we obtain that ${\bf A}$ must be unitary. This is the case commonly used in current FDN practice.

The following theorem gives a general characterization of lossless scattering:
Theorem 1: A scattering matrix (FDN feedback matrix) ${\bf A}$ is lossless if and only if its eigenvalues lie on the unit circle and it admits a basis of linearly independent eigenvectors.
Proof:
In general, the Cholesky factorization ${\bf\Gamma}= {\bf U}^\ast {\bf U}$ gives an upper triangular matrix ${\bf U}$ which converts ${\bf A}$ to a unitary matrix via similarity transformation: ${\bf A}^\ast {\bf\Gamma}{\bf A}= {\bf\Gamma}\Rightarrow {\bf A}^\ast {\bf U}^\... ... {\bf U}^\ast {\bf U}\Rightarrow {\tilde {\bf A}}^\ast {\tilde {\bf A}}={\bf I}$, where ${\tilde {\bf A}}={\bf U}{\bf A}{\bf U}^{-1}$. Hence, the eigenvalues of every lossless scattering matrix lie on the unit circle. It readily follows from similarity to ${\tilde {\bf A}}$ that ${\bf A}$ admits $N$ linearly independent eigenvectors. In fact, ${\tilde {\bf A}}$ is a normal matrix (since it is unitary), and normal matrices admit a basis of linearly independent eigenvectors [21].

Conversely, assume $\vert\lambda\vert = 1$ for each eigenvalue of ${\bf A}$, and that there exists a matrix ${\bf T}$ of linearly independent eigenvectors of ${\bf A}$. Then the matrix ${\bf T}$ diagonalizes ${\bf A}$ to give ${\bf T}^{-1}{\bf A}{\bf T}= {\bf D}\Rightarrow {\bf T}^\ast {\bf A}^\ast {{\bf T}^{-1}}^\ast = {\bf D}^\ast$, where ${\bf D}= \mbox{diag}(\lambda_1,\dots,\lambda_N)$. Multiplying, we obtain ${\bf T}^\ast {\bf A}^\ast {{\bf T}^{-1}}^\ast {\bf T}^{-1}{\bf A}{\bf T}={\bf D... ...}^\ast {{\bf T}^{-1}}^\ast {\bf T}^{-1}{\bf A}={{\bf T}^{-1}}^\ast {\bf T}^{-1}$. Thus, (27) is satisfied for ${\bf\Gamma}={{\bf T}^{-1}}^\ast {\bf T}^{-1}$ which is Hermitian and positive definite. $\Box$

Thus, lossless scattering matrices may be fully parametrized as ${\bf A}= {\bf T}^{-1}{\bf D}{\bf T}$, where ${\bf D}$ is any unit-modulus diagonal matrix, and ${\bf T}$ is any invertible matrix. In the real case, we have ${\bf D}= \mbox{diag}(\pm 1)$ and ${\bf T}\in\Re^{N\times N}$.